In the circuit shown in the figure the transistors M1 and M2 are operating in saturation ...

The given circuit contains two NMOS transistors \(M_1\) and \(M_2\). Find the voltage gain:

\[ A_v=\frac{v_{out}}{v_{in}} \]

Solution

Step 1: Identify the amplifier

The input signal is applied to the gate of \(M_2\). Therefore \(M_2\) is the amplifying transistor.

A MOS transistor converts voltage into current:

\[ i_d=g_m v_{gs} \] For \(M_2\): \[ i_2=g_{m2}v_{in} \]

Step 2: Output resistance of the circuit

For small signal analysis, the power supply \(V_{DD}\) becomes AC ground.

AC Ground

      |
      \(r_{o2}\)
      |
Vout
      |
      \(r_{o1}\)
      |
Ground

Both \(r_{o1}\) and \(r_{o2}\) are connected between the same two nodes. Therefore they are parallel.

\[ R_{out}=r_{o1}\parallel r_{o2} \]

Step 3: Why \(g_{m1}\) is not present

For \(M_1\):

\[ v_{gs1}=0 \] Therefore the transconductance current is: \[ i_{gm1}=g_{m1}v_{gs1} \] \[ i_{gm1}=g_{m1}(0)=0 \]

So \(M_1\) does not contribute a \(g_m\) current. It only contributes its output resistance \(r_{o1}\).

Step 4: Calculate output voltage

The current generated by \(M_2\) is:

\[ i_o=g_{m2}v_{in} \] The output voltage is: \[ v_{out}=-i_oR_{out} \] Substitute: \[ v_{out} = -(g_{m2}v_{in}) (r_{o1}\parallel r_{o2}) \]

Step 5: Voltage gain

Divide by \(v_{in}\): \[ \frac{v_{out}}{v_{in}} = -g_{m2}(r_{o1}\parallel r_{o2}) \]

Important Concepts

• The transistor receiving the input contributes \(g_m\).
• The transistors connected at the output contribute \(r_o\).
• Parallel resistances appear because both resistors connect from output to AC ground.

Final Answer:

\[ \boxed{ A_v= -g_{m2}(r_{o1}\parallel r_{o2}) } \]

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